Exercise 18, § 4

Let $A$ be a noetherian integral domain, $x$ a nonzero element of $A$. Suppose that $A$ is separated and complete for the $xA$-adic topology and that for every prime ideal $p \in \mathrm{Ass}_A(A/xA)$, the ring $A/p$ is Japanese. Prove that $A$ is Japanese. (Let $\overline{A}$ be the integral closure of $A$ in a finite extension of its field of fractions. a) Since the ring $\overline{A}$ is Krull (p. 81, Exercise 14), let $\mathfrak{P}_1, ..., \mathfrak{P}_n$ be the prime ideals of height 1 of $\overline{A}$ containing $x$. b) Let $p_i = \mathfrak{P}_i \cap A$. Then one has $p_i \in \mathrm{Ass}_A(A/xA)$ and $\kappa(\mathfrak{P}_i)$ is a finite extension of $\kappa(p_i)$. (Use Exercises 11 and 13, p. 80.) c) $\overline{A}/\mathfrak{P}_i$ is a finitely generated $A/p_i$-module. d) $\overline{A}/xA$ is a finitely generated $A/xA$-module. e) $A$ is separated for the $xA$-adic topology. Conclude as in the proof of Theorem 1 of § 4, No. 2.)